As defined in (1), (2) or (11), the "hamiltonian" are dimensionless. But a true hamiltonian, representing energy must have a energy dimension. So you need a energy scale $E$,depending on the physical problem, to define the true hamiltonian, for instance $\tilde H_0 = E ~ H_0 $. Maybe the author use these dimensionless hamiltonian to simplify the calculus.
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TrimokJul 16 '13 at 18:11

@Trimok, your comment partially answers my question. Is this kind of dimensionless Hamiltonian show up often in quantum mechanics or quantum information?
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Omar ShehabJul 16 '13 at 19:08

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I don't know exactly. If $E$ is a constant, working with dimensionless hamiltonians is the same thing as working in units $E=1$. For instance, in Quantum field Theory, one very often work with units $c = \hbar = 1$. This simplifyes notations and calculus. As long as you know what $E$ exactly is (and it depends on the particular physical system your are studying), this is not a problem.
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TrimokJul 17 '13 at 7:16

@Trimok, so in QFT you do the calculation for $c = \hbar = 1$ and rescale the final result, right?
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Omar ShehabJul 17 '13 at 12:54

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If you need to do that, yes. For instance, in QED, a lot of results (scattering amplitudes, etcc.) depends only on the coupling constant $\alpha = \frac{e^2}{\hbar c}$. Moreover, $c=1$ allows to put at the same level time, the $x_0$ coordinate and space, $x_1,x_2, x_3$ coordinates - This is Minkowski space-time. Note, in fact, that putting $c=1, \hbar=1, G=1$ is a natural way of thinking, because it corresponds to Planck units, and this simplifyes equations,see Wikipedia.
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TrimokJul 17 '13 at 16:09